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The d in derivatives

  1. Nov 2, 2004 #1
    Does dx or d(any variable) mean an infitesimally change in x (or another variable)?

    If you have dy/dx, can you separate dy from dx? How do you do this?
     
  2. jcsd
  3. Nov 2, 2004 #2
    As fare as I understand it is dy and dx the change on the tangent line. But that you may already know. And, yes, they're infintesimals.

    dx is independent and thus dx = [itex]\Delta x[/itex], but dy != [itex]\Delta y[/itex].
    [tex]dy = f'(x)dx[/tex] or [tex]\frac{dy}{dx}=f'(x)[/tex]

    d(x) is an alternativ notation for f'(x), i think.
     
    Last edited: Nov 2, 2004
  4. Nov 2, 2004 #3

    arildno

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    I'll choose to answer this in this thread; I see you have a similar thread elsewhere:
    Basically, the reason why we occasionally may treat dy/dx analogously to a fraction, in particular, splitting it, is THE CHAIN RULE OF DIFFERENTIATION!
    Suppose you have a differential equation on the form:
    [tex]f(y(x))\frac{dy}{dx}=g(x)[/tex]
    Note the explicit dependendce of x in the composite function f!!
    Assume there exist a function F, so that its derivative with respect to its sole variable is the function f.
    Hence, we may rewrite the lefthand-side in the upper equation, USING THE CHAIN RULE:
    [tex]\frac{d}{dx}F(y(x))=g(x)[/tex]
    Now, let's integrate this equation WITH RESPECT TO x!!
    By the fundamental theorem of calculus, we gain:
    [tex]F(y(x_{1}))-F(y(x_{0}))=\int_{x_{0}}^{x_{1}}g(x)dx (1)[/tex]
    Let us now consider the smart trick:
    [tex]y_{1}=y(x_{1}),y_{0}=y(x_{0})[/tex]
    (1) may therefore be written as:
    [tex]\int_{y_{0}}^{y_{1}}f(y)dy=\int_{x_{0}}^{x_{1}}g(x)dx (2)[/tex]
    since the left-hand side equals, by fund. theorem of calculus:
    [tex]\int_{y_{0}}^{y_{1}}f(y)dy=F(y_{1})-F(y_{0})[/tex]

    But (2) is the way you've been taught to integrate your original differential equation.
     
  5. Nov 2, 2004 #4
    Is y(x) the 'sole variable' of F?

    What happened to [tex]\frac{dy}{dx}[/tex] in [tex]\frac{d}{dx}F(y(x))=g(x)[/tex]?
     
  6. Nov 2, 2004 #5

    arildno

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    By the chain rule:
    [tex]\frac{d}{dx}F(y(x))=F'(y(x))\frac{dy}{dx}=f(y(x))\frac{dy}{dx}[/tex]
    since F'=f
     
  7. Nov 2, 2004 #6
    I think I understand now. It's just hard to get my head around (I guess I'm a little stupid).
     
  8. Nov 2, 2004 #7

    arildno

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    Inexperience is often mistaken for stupidity..:wink:

    Besides, if someone has been taught formulaic maths, rather than the logic and reasons behind formulae, they will often encounter problems generated more by the dubious techniques they've been exposed to, rather than by personal stupidity..:smile:
     
  9. Nov 2, 2004 #8
    Would the best way to ensure that I understand everything be to do as many questions as I possibly can?
     
  10. Nov 2, 2004 #9

    arildno

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    Mmm..exercises are certainly important, indeed crucial, in developing a mathematical skill with your "hand". (That is, getting used to it, building up problem-solving routines and so on).

    However, you should from time to time stop and ask yourself:
    "Do I understand this? Why must it be so?"
    Such sessions, when successfully resolved, while not productive in the sense of numbers of exercises solved, are however, in the long run, the most satisfying experiences in your education.
     
  11. Nov 2, 2004 #10
    That does makes sense.
     
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